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Wave Optics — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsWave Optics2 Marks
Question
In Young's experiment interference bands were produced on a screen placed at \(150 cm\) from two slits, \(0.15 mm\) apart and illuminated by the light of wavelength 6500Å . Calculate the fringe width.
✓

Answer

Given:
$D = 150 cm = 1.5 m,$
$d = 0.15 mm = 1.5 x 10^{-4} m,$
$\lambda = 6500  Å = 6.5 x 10^{-7}m$
To find:
Fringe width ( X )
Formula:
\(X=\frac{\lambda D}{d}\)
Calculation:
From formula,
\(X=\frac{6.5 \times 10^{-7} \times 1.5}{1.5 \times 10^4}\)
\(X=6.5 \times 10^{-3} m\)
$X = 6.5 mm$
The fringe width is 6.5 mm
 

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